What is the meaning that a ring has nontrivial idempotents?

Question

If $\,{\rm char}(F) \ne 2\ $ then $\,F[x]/(x^2-1)\, \cong\, F[x]/(x-1) \oplus F[x]/(x+1)\, \cong\, F^2$ has nontrivial idempotents, e.g. $\,(0,1),\,$ but $F[x]/(x^2)$ does not (as one easily verifies).

I do not understand the sentence that it has nontrivial idempotents (from this answer).

Answer 1

An idempotent is an element $r$ such that $r^2=r$ (idem is latin for self, and potent for power. So an idempotent has itself as a power.)

In any ring with unit, the elements $0$ and $1$ are idempotents (just like they are in $\mathbb{Z}$). So to say a ring has non-trivial idempotents is saying that it has idempotents besides the obvious ones.

In $R = F[x]/(x^2-1)$, let $y=x+1$ (more precisely, the equivalence class of $x+1$ modulo multiples of $x^2-1$). Then $$ y^2 = (x+1)^2 = x^2 + 2x + 1 = 2x+2 = 2(x+1) = 2y $$ This isn't quite an idempotent since we got a multiple of $y$ instead of just $y$. But remember the assumption that $\operatorname{char} F \neq 2$. This means we can divide by $2$.

Let $u = \frac{y}{2}$. Then $$ u^2 = \frac{y^2}{4} = \frac{2y}{4} = \frac{y}{2} = u $$ so $u = \frac{x+1}{2}$ is an idempotent.

As JonathanZ points out in the comments, $\frac{x-1}{-2} = \frac{1-x}{2}$ is another idempotent, for the same reason.

Answer 2

In general, if you have a ring of the form $R_1\times R_2$ (note that identity here is $(1,1)$), then $e=(1,0)$ and $f = (0,1)$ have the property that $e^2=e$ and $f^2=f$. These elements are called idempotent. Note that, in any ring $R$, the elements $0,1\in R$ are idempotent. Those two are called the trivial idempotents. Any other idempotent is called non-trivial.

Answer 3

Following up on MatthewL's correct answer:

Let's let $0$ and $1$ denote the additive and multiplicative identities in your field $F$. Then in the ring $F \times F$ (which is what they mean by $F^2$) $$(0,0)$$ is the additive identity and $$(1,1)$$ is the multiplicative identity (and both are idempotents). But, as operations in $F\times F$ are component by components it's easy to see that both $$ (1,0) \space \text{and} \space(0,1)$$ are also idempotents, and these are the "non-trivial" idempotents they are referring to.

Notice that we got these non-trivial idempotents using only the very weak assumption that our ring can be written as the cross product of two other rings, so it's a pretty general result.